Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{x^2 - 7x + 10}{x^2 - 9x - 10} \div \dfrac{x - 5}{-3x - 3} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{x^2 - 7x + 10}{x^2 - 9x - 10} \times \dfrac{-3x - 3}{x - 5} $ First factor out any common factors. $r = \dfrac{x^2 - 7x + 10}{x^2 - 9x - 10} \times \dfrac{-3(x + 1)}{x - 5} $ Then factor the quadratic expressions. $r = \dfrac {(x - 5)(x - 2)} {(x + 1)(x - 10)} \times \dfrac {-3(x + 1)} {x - 5} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { (x - 5)(x - 2) \times -3(x + 1)} { (x + 1)(x - 10) \times (x - 5)} $ $r = \dfrac {-3(x - 5)(x - 2)(x + 1)} {(x + 1)(x - 10)(x - 5)} $ Notice that $(x + 1)$ and $(x - 5)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-3(x - 5)(x - 2)\cancel{(x + 1)}} {\cancel{(x + 1)}(x - 10)(x - 5)} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $r = \dfrac {-3\cancel{(x - 5)}(x - 2)\cancel{(x + 1)}} {\cancel{(x + 1)}(x - 10)\cancel{(x - 5)}} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $r = \dfrac {-3(x - 2)} {x - 10} $ $ r = \dfrac{-3(x - 2)}{x - 10}; x \neq -1; x \neq 5 $